Why the orthogonal projection $Pu$ is a positive operator if $U$ is a subspace of $V$? To prove it, we need to prove that $Pu$ is both self-adjoint and $<Puv,v>\ge 0$ for all $v∈V$. It is obvious that $<Puv,v>\ge0$, but why is it self-adjoint?
 A: Self-adjoint is equivalent to the existence of an orthonormal basis $B$, such that the matrix with respect to $B$ is symmetric (With respect to an orthonormal basis, taking the adjoint operator is the same as taking the transpose of the matrix).
Take an orthonormal basis of $U$ and an orthonormal basis of $U^\perp$. Together, they form an orthonormal basis of $V$ and the matrix with respect to this basis is given by the symmetric matrix
$$\begin{pmatrix}I_r & 0 \\0&0\end{pmatrix}$$, where $I_r$ is the $r \times r$-unit matrix. (r is the dimension of $U$)
As an alternative, you can just test the equality
$$\langle Px,y \rangle = \langle x,Py \rangle$$
when $x,y$ run through the orthonormal basis above.
A: In order to prove that $P$ is self-adjoint (without selecting a basis), we should prove that for any vectors $x,y$:
$$
\langle Px,y \rangle = \langle x,Py \rangle
$$
where the brackets denote the inner product (that is, the "dot-product").  Note that each vector can be written as the sum of perpendicular components, with one in $U$. We can write
$$
\langle Px,y\rangle = \\
\langle P(x^{||}+x^\perp),y^{||} + y^\perp \rangle = \\
\langle x^{||},y^{||} + y^\perp \rangle = \\
\langle x^{||},y^{||} \rangle + 
\langle x^{||},y^\perp \rangle =\\ 
\langle x^{||},y^{||} \rangle + 
0 = \\
\langle x^{||},y^{||} \rangle + 
\langle x^{\perp}, y^{||} \rangle = \\
\langle x^{||}+x^{\perp},y^{||} \rangle = \\
\langle x,Py \rangle
$$
